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Question
In the given figure, bisector of ∠BAC intersects side BC at point D. Prove that AB > BD.
Solution
Given: Bisector of ∠BAC intersects side BC at point D.
To prove: AB > BD
Proof:
∠BAD ≅ ∠DAC ...(i) ...[Seg AD bisects ∠BAC]
∠ADB is the exterior angle of ∆ADC.
∴ ∠ADB > ∠DAC ...(ii) ...[Property of exterior angle]
∴ ∠ADB > ∠BAD ...(iii) ...[From (i) and (ii)]
In ΔABD,
∠ADB > ∠BAD ...[From (iii)]
∴ AB > BD ...[Side opposite to greater angle is greater]
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